Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx-9$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-8$$ $$ab=1\left(-9\right)=-9$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-9$.
$$1,-9$$ $$3,-3$$
Calculate the sum for each pair.
$$1-9=-8$$ $$3-3=0$$
The solution is the pair that gives sum $-8$.
$$a=-9$$ $$b=1$$
Rewrite $x^{2}-8x-9$ as $\left(x^{2}-9x\right)+\left(x-9\right)$.
$$\left(x^{2}-9x\right)+\left(x-9\right)$$
Factor out $x$ in $x^{2}-9x$.
$$x\left(x-9\right)+x-9$$
Factor out common term $x-9$ by using distributive property.
$$\left(x-9\right)\left(x+1\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$x^{2}-8x-9=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
